Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit problem

  1. Sep 30, 2004 #1
    i did this problem but i just wanted some confirmation if im right.....

    lim (3+h)^-1 - (3)^-1 / h
    h->0


    i worked it out...but still got 0....maybe some confirmation and if u could show ur work that would be nice too......thx in advance



    ps my first post!!!
     
  2. jcsd
  3. Sep 30, 2004 #2

    T@P

    User Avatar

    if im not mistaken thats a derivative in one of it (many) forms.
    basically, it x' at the point 3... which would be 1 if i understood your notation right. (is the h under the whole fractoin or just the end?)

    hope im not confusing you too much.
     
  4. Oct 1, 2004 #3
    Based on what you are saying the function in question is a constant function. Meaning the function is of the form y=1/3, which is just a boring horizontal line. Now consider what a derivative is suppose to be. A derivative gives information about a functions slope at values of x in the domain of the function. Now consider your horizontal line, what can you tell me about its slope?

    Best Regards
     
  5. Oct 1, 2004 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor


    How did you get that interpretation?

    Assuming the orginal post was lim ((3+h)-1- 3-1)/h
    (notice the additional parentheses) then this is the derivative of x-1 evaluated at x= 3. Assuming that the purpose of this is to actually calculate that derivative (so that you can't just use the derivative itself to get the limit!) then the best way to do it is to combine the fractions:

    [tex]\frac{1}{3+h}- \frac{1}{3}= \frac{3- (3+h)}{3(3+h)}=\frac{-h}{3(3+h)}[/itex]
    so the "difference quotient" becomes
    [tex]\frac{-h}{3h(3+h)}[/itex].
    As long as h is not 0 that is the same as
    [tex]\frac{-1}{3(3+h)}[/itex]
    and it should be easy to find the limit as h goes to 0.
     
  6. Oct 1, 2004 #5

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    I got zero on the work as well. Find the limit.

    [tex]\frac{\frac{-1}{3(3+h)}}{h}[/tex]

    Simplify and use direct substitution.

    I got zero.

    I could be wrong though.

    [tex]\frac{\frac{-1}{3(3+h)}}{h}[/tex]

    to

    [tex]\frac{-h}{3(3+h)}[/tex]

    Now, substitute h~0.

    Note: I excluded lim in my work for simplicities sake.
     
    Last edited: Oct 1, 2004
  7. Oct 1, 2004 #6

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Also, I'm a Brock University student. ;)
     
  8. Oct 1, 2004 #7

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Jason,

    You dropped a factor of h in the numerator. Halls' analysis is correct.
     
  9. Oct 1, 2004 #8

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    We aren't looking for the rate of change at x=3.

    We are looking for the limit.

    The limit of 1/x is 0.
     
  10. Oct 2, 2004 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Brock University may get upset at you for using their name!

    In the first place, "the limit of 1/x is 0" is meaningless- you have to say "limit of 1/x" as x goes to some specific value4. The only number that would give a limit of 0 for 1/x is infinity and infinity has nothing to do with the original problem.

    "We aren't looking for the rate of change at x=3."

    Perhaps you aren't but anyone who is trying to answer the original question is!
     
  11. Oct 2, 2004 #10

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    He/she wasn't looking for that. I am in the same program because we have the same assignment. The two questions she asked came from the same school.

    About the limit mistake, there's an even bigger one. I thought about it last night while going to bed, and thought about what I said. First, it didn't make any sense, like you explained. Second, and I can't believe you didn't spot this, the limit doesn't really exist because the left hand limit doesn't equal the right hand limit.

    Note: I have every right to say I'm a Brock student.
     
  12. Oct 2, 2004 #11

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Just so you know the assignments have been handed in.
     
  13. Oct 2, 2004 #12

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Well then be sure to report back to us when you find the "right" answer! :smile:
     
  14. Oct 2, 2004 #13

    T@P

    User Avatar

    whoa
    why doesnt the original poster just re-state the question so there's no more ambiguity?
    I personally am not sure wether the last h refers to the whole limit or not...
     
  15. Oct 3, 2004 #14
    I have no idea why I thought what I did but I am glad you were able to correct it. Sorry about that. Next time I try to help I will be a lot more careful before posting.

    Regards
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Limit problem
  1. Limit problem. (Replies: 19)

  2. A limit problem (Replies: 9)

  3. Limit problem (Replies: 2)

  4. Limit Problem (Replies: 5)

  5. Problem with limit (Replies: 4)

Loading...